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Math Help - Prove Convergence

  1. #1
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    Prove Convergence

    1a.) Prove the sequence converges which is defined by x_1 = 3 and x_(n + 1) = 1/(4 - x_n)

    b.) Now we know lim x_n exists... explain why lim x_(n + 1) must also exist, and further, explain why it must equal the same value.

    c.) Of the recursive equation shown in a.), take the lim of each side of the recursive eq'n. to explicity compute lim x_n

    My work:

    a.) Now I am not exactly sure how to prove this... I've learned the epsilon method, showing that there exists one greater than 0.. blah blah blah. But I was thinking of maybe trying to use proof by induction to show:

    0 <= x_n <= 3 and x_(n) is decreases, which will imply that {x_n} converges (by the monotone convergence thm)...

    b.) Not sure..

    c.) Maybe something like:

    x = 1/(4 - x) => x = 2 - sqrt(3) since 0 <= x <= 3

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    Help?
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by seerTneerGevoLI View Post
    1a.) Prove the sequence converges which is defined by x_1 = 3 and x_(n + 1) = 1/(4 - x_n)

    b.) Now we know lim x_n exists... explain why lim x_(n + 1) must also exist, and further, explain why it must equal the same value.

    c.) Of the recursive equation shown in a.), take the lim of each side of the recursive eq'n. to explicity compute lim x_n

    My work:

    a.) Now I am not exactly sure how to prove this... I've learned the epsilon method, showing that there exists one greater than 0.. blah blah blah. But I was thinking of maybe trying to use proof by induction to show:

    0 <= x_n <= 3 and x_(n) is decreases, which will imply that {x_n} converges (by the monotone convergence thm)...
    Yes

    b.) Not sure..
    Look at the epsilon method and assume that you have the N_{\varepsilon} for the sequence x_n, then this will also do for the sequence x_{n+1}

    c.) Maybe something like:

    x = 1/(4 - x) => x = 2 - sqrt(3) since 0 <= x <= 3
    Yes

    RonL
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  3. #3
    MHF Contributor red_dog's Avatar
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    Quote Originally Posted by seerTneerGevoLI View Post
    b.) Now we know lim x_n exists... explain why lim x_(n + 1) must also exist, and further, explain why it must equal the same value.
    If  (x_n)_{n\in\mathbf{N}} is a sequence and \lim_{n\to\infty}x_n=l, then any subsequence of the sequence has the same limit.

    (x_{n+1})_{n\in\mathbf{N}} is a subsequence of  (x_n)_{n\in\mathbf{N}}.
     (x_n)_{n\in\mathbf{N}} is convergent, so (x_{n+1})_{n\in\mathbf{N}} is also convergent to the same limit.
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